From: Dr. Sydney
Date: Wed, 9 Nov 1994 11:40:33 -0500 (EST)
Dear Jessica,
Thanks for writing Dr. Math! Your question about exponents is a
good question. Before we look at any of the specific examples you
wanted help with, let's look at the theory behind it all:
We know (x^a)(x^b)=x^(a+b), right? This is because x^a is x
multiplied by itself a times and x^b is x multiplied by itself b times, and
if we take the product of these two we get x multiplied by itself (a+b)
times. Let's look at an example with real numbers:
What is (x^2)(x^3)=?
Well, x^2 =(x)(x) and x^3=(x)(x)(x), so
(x^2)(x^3)=(x)(x)(x)(x)(x)=x^5
So, (x^2)(x^3)=x^(2+3)=x^5
The other thing to remember when doing problems like the ones you
are doing is that
b ab
(x^a) = x
This is because
b
(x^a) is just x^a multiplied by itself b times, and since x^a is
x multiplied by itslef a times,
b
(x^a) is x multiplied by itself ab times.
Let's look at an example of this:
What is
2
(x^3) = ? Well, 2
3
x is x^3 multiplied by itself 2 times, so
2
(x^3) = (x^3)(x^3). From above, though we know that
(x^3)(x^3) = x^(3+3) = x^6 = x ^[(3)(2)]
So,
2 (3)(2)
(x^3) = x
Okay, so now we can do those problems you were asking about. I'll
work through one of them with you, and see if you can do the rest now
that you've seen how these exponents work:
1) (2x^2 y^3)^4(-x y^2)^2 = ?
All we need to do for this problem is use the properties of exponents
that we went over above:
well, (2x^2 y^3)^4 = (2^4)(x^8)(y^12) = 16(x^8)(y^12), right?
and (-x y^2)^2 = (-x)^2(y^4)=(x^2)(y^4)
So, our answer will be:
16(x^8)(y^12)(x^2)(y^4)=16(x^(8+2))(y^(12+4))=16(x^10)(y^16)
That is as reduced as it will get. Do all of these steps make sense to
you? If you have any questions about this or any other problems,
please do feel free to write back. Thanks for writing!
--Sydney, Dr. "exponent"